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A063037
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Numbers without 3 consecutive equal binary digits.
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8
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0, 1, 2, 3, 4, 5, 6, 9, 10, 11, 12, 13, 18, 19, 20, 21, 22, 25, 26, 27, 36, 37, 38, 41, 42, 43, 44, 45, 50, 51, 52, 53, 54, 73, 74, 75, 76, 77, 82, 83, 84, 85, 86, 89, 90, 91, 100, 101, 102, 105, 106, 107, 108, 109, 146, 147, 148, 149, 150, 153, 154, 155, 164, 165, 166
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,3
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COMMENTS
| Complement of A136037; intersection of A003796 and A003726. - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Dec 11 2007
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LINKS
| R. Zumkeller, Table of n, a(n) for n = 1..10000
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FORMULA
| It appears (but has not as yet been proved) that the terms of {a(n)} can be computed recursively as follows. Let {c(i)} be defined for i>=4 by c(i) = 2c(i-1) + 1, if n is a multiple of 3, else c(i) = 2c(i-1) - 1, with c(4) = 1. I.e., {c(i)}={1,1,3,5,9,19,37,73,147,...}, for i=4,5,6,... . Let a(1)=1, a(2)=2, a(3)=3. For n>3, choose k so that F(k)-2<n<=F(k+1)-2, where F(k) denotes the k-th Fibonacci number (A000045). Then a(n)=c(k)+2a(F(k)-2)-a(2F(k)-n-3). This has been verified for n up to 1100. [From John W. Layman (layman(AT)math.vt.edu), May 26 2009]
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EXAMPLE
| The binary representation of 9 (1001) has no 3 consecutive equal digits.
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PROG
| (PARI) { n=0; for (m=0, 10^9, x=m; t=1; b=2; while (x>0, d=x-2*(x\2); x\=2; if (d==b, c++; if (c==3, t=x=0), b=d; c=1)); if (t, write("b063037.txt", n++, " ", m); if (n==1000, break)) ) } [From Harry J. Smith (hjsmithh(AT)sbcglobal.net), Aug 16 2009]
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CROSSREFS
| Cf. A000975.
A000045 [From John W. Layman (layman(AT)math.vt.edu), May 26 2009]
Sequence in context: A032878 A032845 A023776 * A201992 A157189 A136250
Adjacent sequences: A063034 A063035 A063036 * A063038 A063039 A063040
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KEYWORD
| easy,nonn
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AUTHOR
| Lior Manor (lior.manor(AT)gmail.com) Jul 05 2001
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EXTENSIONS
| Missing "less than" sign supplied in the conjectured recurrence; thanks to Franklin T. Adams-Watters for pointing this out John W. Layman (layman(AT)math.vt.edu), Nov 09 2009
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